(4) \( \frac{2}{3}+\frac{3 x+2}{x}-\frac{x+4}{2 x} \)
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To combine the fractions \( \frac{2}{3}+\frac{3 x+2}{x}-\frac{x+4}{2 x} \), we need a common denominator. The denominators here are \( 3, x, \) and \( 2x \). The least common multiple of these is \( 6x \). Now, we can rewrite each fraction with the common denominator: 1. \( \frac{2}{3} = \frac{2 \cdot 2x}{3 \cdot 2x} = \frac{4x}{6x} \) 2. \( \frac{3x+2}{x} = \frac{(3x+2) \cdot 6}{x \cdot 6} = \frac{18x + 12}{6x} \) 3. \( \frac{x+4}{2x} = \frac{(x+4) \cdot 3}{2x \cdot 3} = \frac{3x + 12}{6x} \) Now, substitute these back into the equation: \[ \frac{4x}{6x} + \frac{18x + 12}{6x} - \frac{3x + 12}{6x} \] Combine all the numerators over the common denominator: \[ \frac{4x + 18x + 12 - (3x + 12)}{6x} \] This simplifies to: \[ \frac{4x + 18x + 12 - 3x - 12}{6x} = \frac{19x}{6x} \] Finally, simplify: \[ \frac{19}{6} \] So, the final result is: \[ \frac{19}{6} \]